Differentiating products of two functions using the product rule.
Answer at least 3 of 3 correctly to complete this section.
You know how to differentiate x3, sinx, e2x, and many other functions individually. But what if you need to differentiate x3sinx? Or x2e2x?
These are products — two separate functions multiplied together. You cannot simply differentiate each part and multiply the results. That gives the wrong answer every time.
The product rule is the correct tool. It tells you exactly how to handle the derivative of a product, and it appears in almost every A2 differentiation question. Combined with the chain rule, it unlocks the vast majority of functions you will meet at A-Level.
The product rule applies when you have two functions of x multiplied together:
y=first functionf(x)×second functiong(x)
Here are some examples:
| Expression | First function u | Second function v | Product rule needed? |
|---|---|---|---|
| x2sinx | x2 | sinx | Yes |
| x3e2x | x3 | e2x | Yes |
| (2x+1)cosx | 2x+1 | cosx | Yes |
| 3x2(x+5) | 3x2 | x+5 | No — just expand to 3x3+15x2 |
| sin(x2) | — | — | No — this is a composite function (chain rule) |
Key question: Are there two different types of function being multiplied? If one is a polynomial and the other is trig or exponential, you almost certainly need the product rule. If both are polynomials, it is usually easier to expand first.
These two rules handle fundamentally different structures:
| Structure | Example | Rule |
|---|---|---|
| f(x)×g(x) — a product | x2sinx | Product rule |
| f(g(x)) — a composition | sin(x2) | Chain rule |
Notice the difference: x2sinx is ”x2 times sinx”, but sin(x2) is ”sin of x2”. They look similar but have completely different derivatives.
Watch out: Many A2 questions require both rules in the same problem. For example, differentiating x3e2x uses the product rule for the overall structure and the chain rule to differentiate e2x.
Answer at least 3 of 4 correctly to complete this section.
If y=uv, where u and v are both functions of x, then:
dxdy=udxdv+vdxdu
Or in function notation: if y=f(x)g(x), then:
dxdy=f(x)g′(x)+f′(x)g(x)
Think about what happens when both u and v change by small amounts. If u increases by δu and v increases by δv, then the product changes by:
(u+δu)(v+δv)−uv=uδv+vδu+δuδv
The last term (δu⋅δv) is negligibly small compared to the other two, so the rate of change is:
dxdy=udxdv+vdxdu
Memory aid: “Keep the first, differentiate the second — plus — keep the second, differentiate the first.” Some students remember this as “one d’d, one not — plus — swap”.
Differentiate y=x2sinx
Step 1: Let u=x2 and v=sinx.
Step 2: dxdu=2x and dxdv=cosx.
Step 3: Apply the product rule:
dxdy=x2cosx+2xsinx
That is the final answer. You could factorise out x if you wish:
dxdy=x(xcosx+2sinx)
Differentiate y=x3e2x
Let u=x3 and v=e2x.
dxdu=3x2,dxdv=2e2x(chain rule!)
dxdy=x3⋅2e2x+e2x⋅3x2=2x3e2x+3x2e2x
Factorise: dxdy=x2e2x(2x+3)
Watch out: When differentiating e2x, the chain rule gives 2e2x, not just e2x. Forgetting this factor of 2 is a very common error in product rule questions.
Differentiate y=exsinx
Let u=ex and v=sinx.
dxdu=ex,dxdv=cosx
dxdy=excosx+exsinx=ex(cosx+sinx)
Differentiate y=(3x+1)cosx
Let u=3x+1 and v=cosx.
dxdu=3,dxdv=−sinx
dxdy=(3x+1)(−sinx)+cosx⋅3=−(3x+1)sinx+3cosx
Answer at least 3 of 4 correctly to complete this section.
In many A2 questions, you need the product rule for the overall structure and the chain rule to differentiate one (or both) of the factors.
Differentiate y=x2cos(3x)
Let u=x2 and v=cos(3x).
dxdu=2x
For dxdv, we need the chain rule on cos(3x):
dxdv=−3sin(3x)
Now apply the product rule:
dxdy=x2⋅(−3sin(3x))+cos(3x)⋅2x
dxdy=−3x2sin(3x)+2xcos(3x)
Factorising: dxdy=x[−3xsin(3x)+2cos(3x)]
Differentiate y=e2xsin(5x)
Let u=e2x and v=sin(5x).
dxdu=2e2x(chain rule),dxdv=5cos(5x)(chain rule)
dxdy=e2x⋅5cos(5x)+sin(5x)⋅2e2x
dxdy=e2x[5cos(5x)+2sin(5x)]
Exam tip: When factorising product rule answers, look for common factors of e(⋅), x, or trig functions. Examiners often expect a factorised final answer.
This extension is rarely tested and goes beyond most exam board specifications. You do not need it for standard A-Level questions, but it is here if you are curious.
Occasionally you meet a product of three functions. Extend the rule by grouping:
Differentiate y=x2exsinx
Group as y=(x2)(exsinx). Let u=x2 and v=exsinx.
First find dxdv using the product rule on exsinx:
dxdv=excosx+exsinx=ex(cosx+sinx)
Now apply the product rule to the whole thing:
dxdy=x2⋅ex(cosx+sinx)+exsinx⋅2x
dxdy=xex[x(cosx+sinx)+2sinx]
dxdy=xex[xcosx+xsinx+2sinx]
Answer at least 3 of 4 correctly to complete this section.
The curve C has equation y=x2e−x.
(a) Find dxdy.
(b) Find the coordinates of the stationary points of C.
(c) Determine the nature of each stationary point.
(a) Let u=x2, v=e−x.
dxdu=2x,dxdv=−e−x
dxdy=x2(−e−x)+e−x(2x)=−x2e−x+2xe−x=xe−x(2−x)
(b) Set dxdy=0:
xe−x(2−x)=0
Since e−x>0 for all x, the solutions are x=0 or x=2.
At x=0: y=0. Point: (0,0).
At x=2: y=4e−2. Point: (2,4e−2).
(c) We need dx2d2y. Differentiate dxdy=2xe−x−x2e−x using the product rule on each term:
dx2d2y=(2e−x−2xe−x)−(2xe−x−x2e−x)
=2e−x−4xe−x+x2e−x=e−x(x2−4x+2)
At x=0: dx2d2y=e0(0−0+2)=2>0 → local minimum.
At x=2: dx2d2y=e−2(4−8+2)=−2e−2<0 → local maximum.
Find the equation of the tangent to y=xcosx at x=2π.
Step 1: Find dxdy using the product rule.
dxdy=x(−sinx)+cosx⋅1=−xsinx+cosx
Step 2: Evaluate the gradient at x=2π.
dxdy=−2πsin2π+cos2π=−2π(1)+0=−2π
Step 3: Find the y-coordinate: y=2πcos2π=2π⋅0=0.
Step 4: Tangent equation: y−0=−2π(x−2π)
y=−2πx+4π2
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
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